(ii) The Particular Enunciation repeats in special terms the statement already made, and refers it to a diagram, which enables the reader to follow the reasoning more easily. (iii) The Construction then directs the drawing of such straight lines and circles as may be required to effect the purpose of a problem, or to prove the truth of a theorem. (iv) Lastly, the Demonstration proves that the object proposed in a problem has been accomplished, or that the property stated in a theorem is true. Euclid's reasoning is said to be Deductive, because by a connected chain of argument it deduces new truths from truths already proved or admitted. The initial letters Q. E. F., placed at the end of a problem, stand for Quod erat Faciendum, which was to be done. The letters Q. E.D. are appended to a theorem, and stand for Quod erat Demonstrandum, which was to be proved. A Corollary is a statement the truth of which follows readily from an established proposition; it is therefore appended to the proposition as an inference or deduction, which usually requires no further proof. The following symbols and abbreviations may be employed in writing out the propositions of Book I., though their use is not recommended to beginners. and all obvious contractions of words, such as opp., adj., diag., &c., for opposite, adjacent, diagonal, &c. SECTION I. PROPOSITION 1. PROBLEM. To describe an equilateral triangle on a given finite straight line. Let AB be the given straight line. It is required to describe an equilateral triangle on AB. Construction. From centre A, with radius AB, describe the circle BCD. Post. 3. From centre B, with radius BA, describe the circle ACE. Post. 3. From the point C at which the circles cut one another, draw the straight lines CA and CB to the points A and B. Then shall ABC be an equilateral triangle. Post. 1. Because A is the centre of the circle BCD, Proof. Def. 11. And because B is the centre of the circle ACE, Def. 11. But it has been shewn that AC is equal to AB; But things which are equal to the same thing are equal to one another. Therefore AC is equal to BC. and it is described on the given straight line AB. Ax. 1. Q. E. F, PROPOSITION 2. PROBLEM. From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line. It is required to draw from the point A a straight line equal to BC. Construction. Join AB; Post. 1. and on AB describe an equilateral triangle DAB. From centre B, with radius BC, describe the circle CGH. I. 1. Post. 3. Post. 2. Produce DB to meet the circle CGH at G. From centre D, with radius DG, describe the circle GKF. Produce DA to meet the circle GKF at F. Post. 2. Then AF shall be equal to BC. Proof. Because B is the centre of the circle GGH, therefore BC is equal to BG. And because D is the centre of the circle GKF, Def. 11. Def. 11. and DA, DB, parts of them are equal; Def. 19. therefore the remainder AF is equal to the remainder BG. And it has been shewn that BC is equal to BG; Ax. 3. But things which are equal to the same thing are equal to one another. Therefore AF is equal to BC; and it has been drawn from the given point A. Ax. 1. Q. E. F. [This Proposition is rendered necessary by the restriction, tacitly imposed by Euclid, that compasses shall not be used to transfer distances.] PROPOSITION 3. PROBLEM. From the greater of two given straight lines to cut of a part equal to the less. Let AB and C be the two given straight lines, of which AB is the greater. It is required to cut off from AB a part equal to C. Construction. From the point A draw the straight line I. 2. and from centre A, with radius AD, describe the circle DEF, meeting AB at E. Post. 3. Then AE shall be equal to C. Proof. Because A is the centre of the circle DEF, therefore AE is equal to AD. But C is equal to AD. Therefore AE and C are each equal to AD. Therefore AE is equal to C ; and it has been cut off from the given straight line AB. Def. 11. Constr. Q.E.F EXERCISES. 1. On a given straight line describe an isosceles triangle having each of the equal sides equal to a given straight line. 2. On a given base describe an isosceles triangle having each of the equal sides double of the base. 3. In the figure of 1. 2, if AB is equal to BC, shew that D, the vertex of the equilateral triangle, will fall on the circumference of the circle CGH. Obs. Every triangle has six parts, namely its three sides and three angles. Two triangles are said to be equal in all respects, when they can be made to coincide with one another by superposition (see note on Axiom 8), and in this case each part of the one is equal to a corresponding part of the other. PROPOSITION 4. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal; then shall their bases or third sides be equal, and the triangles shall be equal in area, and their remaining angles shall be equal, each to each, namely those to which the equal sides are opposite: that is to say, the triangles shall be equal in all respects. Let ABC, DEF be two triangles, which have the side AB equal to the side DE, the side AC equal to the side DF, and the contained angle BAC equal to the contained angle EDF. Then shall the base BC be equal to the base EF, and the triangle ABC shall be equal to the triangle DEF in area; and the remaining angles shall be equal, each to each, to which the equal sides are opposite, namely the angle ABC to the angle DEF, and the angle ACB to the angle DFE. For if the triangle ABC be applied to the triangle DEF, so that the point A may be on the point D, and the straight line AB along the straight line DE, then because AB is equal to DE, Hyp. therefore the point B must coincide with the point E. |